Singular Procedure: Vanishing polynomial function (A Singular Introduction to Commutative Algebra)

Question

I have recently started studying the computer algebra system Singular with help of the book „A Singular Introduction to Commutative Algebra“ by Gert-Martin Greuel & Gerhard Pfister.

Since a few days now I‘m stuck on Exercise 1.1.13. „Write a SINGULAR procedure, depending on two integers $p,d$, with $p$ prime, which returns all polynomials in $\mathbb{F}_p[X]$ of degree $d$ such that the corresponding polynomial function vanishes. Use the procedure to display all $f \in (\mathbb{Z}/5 \mathbb{Z})[X]$ of degree $\leq$ 6 such that $\tilde{f}=0$“ (Annotation: $\tilde{f}$ is the polynomial function).

If I had to write that function in Python/ C++, I would recursively define all polynomials of degree $d-p$ in $\mathbb{F}_p[X]$ and then multiply all of them with $X^p-X$. Unfortunately that algorithm did not work in Singular as expected. Since this is one of the first exercises of the book I think that there has to be any clue working with Singular functionalities to solve that exercise in a more convenient way. Working through the libraries and Singular functionalities in the book I can‘t find any suitable procedure or functionality that would be helpfull in this case. Either my algorithm is wrong or Singular is requiring a different solution.

Now I‘m requesting help from someone with Singular experience or anyone that studied the book mentioned above.

Answer 1

Before anything else, beware that the desired procedure itself demands exorbitant amounts of runtime: For given values $p$ and $d$, if $d\geq p $ the output consists of $p^{d-p+1}$ polynomials over $\Bbb{F}_p$, requiring a total of $$(d+1)p^{d-p+1},$$ elements of $\Bbb{F}_p$ to represent. As the size of the desired output is exponential in $d$, certainly the runtime will at least exponential.

---

As for an effective algorithm; as you note a polynomial with coefficients in $\Bbb{F}_p$ vanishes on $\Bbb{F}_p$ if and only if it is divisible by $X^p-X$. Polynomial long division shows that mod $X^p-X$ we have $X^i\equiv X^{i+(p-1)}$ for all $i\geq 1$, and so for an arbitrary polynomial $f=\sum_{i=0}^d c_iX^i\in\Bbb{F}_p[X]$ we have $$f\equiv c_0+\sum_{i=1}^{p-1}\left(\sum_{j\geq0}c_{i+(p-1)j}\right)X^i\pmod{X^p-X}.$$ This shows that $f$ vanishes if and only if $c_0=0$ and $\sum_{j\geq0}c_{i+(p-1)j}=0$ for all $i\in\{1,\ldots,p-1\}$.

In particular, if $d\leq p-1$ then this shows that $f\equiv0$, so there are no such polynomials for $d<p$, except perhaps $d=0$ if you consider the zero polynomial to have degree $0$. If $d\geq p$ then the constraints on the coefficients are equivalent to $$c_0=0\qquad\text{ and }\qquad c_i=-\sum_{j\geq1}c_{i+(p-1)j}\quad\text{ for all }\quad i\in\{1,\ldots,p-1\}.$$ In other words, the first $p$ coefficients of $f$ are uniquely determined by the remaining coefficients of $f$, and there are no constraints on the remaining coefficients of $f$. So every choice of coefficients $c_p,c_{p+1},\ldots,c_d\in\Bbb{F}_p$ (with $c_d\neq0$) yields a unique polynomail of degree $d$ that vanishes on $\Bbb{F}_p$. This yields the following algorithm:

1. For $c_p,c_{p+1},\ldots,c_{d-1}$ in $\Bbb{F}_p$ and $c_d\in\Bbb{F}_p^{\times}$:
2. $\qquad$For $i$ in $\{1,\ldots,p-1\}$
3. $\qquad\qquad$Set $c_i:=-\sum_{j\geq1}c_{i+(p-1)j}$.
4. $\qquad$Print $f:=\sum_{i=1}^dc_iX^i$.

This requires $d+2-2p$ additions per polynomial, so the runtime is a linear in the output.

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Another way to reach the same construction is as follows:

1. Take an arbitrary polynomial $f\in\Bbb{F}_p[X]$ of degree $d$.
2. Compute the unique $g\in\Bbb{F}_p[X]$ with $\deg g<p$ such that $g\equiv f\pmod{X^p-X}$.
3. Output $f-g$.

The computation in step $2$ is a matter of polynomial long division. Then $$f=(X^p-X)h+g,$$ which shows that $f-g$ vanishes on $\Bbb{F}_p$. Of course every polynomial of degree $d$ that vanishes on $\Bbb{F}_p$ arises in this way, because if $f\in\Bbb{F}_p[X]$ vanishes on $\Bbb{F}_p$ then $g\equiv0$.

Answer 2

I think I finally did it after uncountable tedious hours! The coding is not very elegant but Singular is a little bit sensitive as far as for-loops, recursion and deceleration is concerned. Due to the recursion this implementation will not work for larger $d-p$

ring A = 5, x, dp;
LIB "general.lib";

proc allpoly(int p, int d, int #) // p prime, d degree
{
    list L;
    int j;
    int k;
    if(d!=0)
    {
        list A = allpoly(p,d-1,0);
        for(int i=1; i<=size(A);i++)
        {
            if(# == 0)
            {
                for(k=0; k<p; k++)
                {
                    L = insert(L,k*x^d + A[i]);
                }
            }
            else
            {
                for(j=1; j<p; j++)
                {
                    L = insert(L,j*x^d + A[i]);
                }
            }
        }
    } 
    else
    {
        for(int t=0;t<p;t++)
        {
            L = insert(L,t);
        }
    }
    return(L);
}

attrib(allpoly,"default_arg",1);

proc multi(poly f) // multiplying a polynomial with x^p - x where p is the characteristic of ring A
{
    return(f * (x^(ringlist(A)[1]) - x));
}

proc vanish(int p, int d) // p prime, d degree, d => p
{
    if(p>d)
    {
        list L = (0);
        return(L);
    }
    else
    {
        list A = allpoly(p,d-p);
        list L = apply(A, multi);
        if(p==d) // will sort out the 0 polynomial if p = d
        {
            L = delete(L,size(L));
        }
        return(L);
    }
}

If we now apply for instance vanish(5,6) we get the output:

[1]:
   -x6+x2
[2]:
   -2x6+2x2
[3]:
   2x6-2x2
[4]:
   x6-x2
[5]:
   -x6+x5+x2-x
[6]:
   -2x6+x5+2x2-x
[7]:
   2x6+x5-2x2-x
[8]:
   x6+x5-x2-x
[9]:
   -x6+2x5+x2-2x
[10]:
   -2x6+2x5+2x2-2x
[11]:
   2x6+2x5-2x2-2x
[12]:
   x6+2x5-x2-2x
[13]:
   -x6-2x5+x2+2x
[14]:
   -2x6-2x5+2x2+2x
[15]:
   2x6-2x5-2x2+2x
[16]:
   x6-2x5-x2+2x
[17]:
   -x6-x5+x2+x
[18]:
   -2x6-x5+2x2+x
[19]:
   2x6-x5-2x2+x
[20]:
   x6-x5-x2+x

which should be correct.

If someone has a better solution that maybe does use some exclusive Singular functionalities which I'm not aware of I would be glad if you could share it with me.